Cumulative Incidence From Cause-Specific Joint Hazards

From Cause-Specific Hazards to Event Probabilities

A cause-specific hazard describes the instantaneous rate of one event type among individuals who are still event-free. A cumulative incidence function (CIF) answers a different, absolute-risk question: what is the probability that a specified event occurs by time $t$, before any competing event? The conversion requires accounting for the fact that a competing event removes an individual from the risk set for the event of interest.

In a joint model, cause-specific hazards may depend on latent random effects that also describe longitudinal trajectories. The CIF is first obtained conditional on those effects; a population-level CIF then averages the conditional probability over their distribution. The order matters: in general, averaging the hazards first and then calculating a CIF does not give the same answer.

Definition: Let $T$ be the time to the first event and $J\in\{1,\ldots,K\}$ its cause. For cause $k$, the cumulative incidence function is $F_k(t)=P(T\leq t,\ J=k)$. Given covariates $x$ and random effects $b$, let $\lambda_k(t\mid b,x)$ be the cause-specific hazard for cause $k$. The conditional event-free survival probability is $S(t\mid b,x)=P(T>t\mid b,x)$. The conditional cause-specific cumulative incidence is $$ F_k(t\mid b,x)=\int_0^t S(u\mid b,x)\lambda_k(u\mid b,x)\,du. $$

This expression assumes that the competing event types are mutually exclusive first events, that the cause-specific hazards correctly describe the event process conditional on $b$ and $x$, and that the hazards are nonnegative and integrable over the time interval considered. It also assumes that the random-effects distribution used for population averaging is appropriate for the target population. Censoring is not a competing event in this definition: censoring prevents observation of an event time but does not itself constitute an event of cause $k$.

Why Event-Free Survival Appears in the Integral

Write $\lambda_{\mathrm{all}}(t\mid b,x)=\sum_{j=1}^K\lambda_j(t\mid b,x)$ for the total hazard of any first event. Over a short interval of length $dt$, conditional on being event-free just before $t$, the probability of a cause-$k$ event is approximately $\lambda_k(t\mid b,x)dt$. The probability of having remained event-free up to that time is $S(t\mid b,x)$. Thus the probability contribution to cause $k$ near time $t$ is approximately $S(t\mid b,x)\lambda_k(t\mid b,x)dt$, and summing those contributions over time gives the integral in the definition.

The survival term is determined by all causes, not only the target cause. Its differential equation is

\frac{d}{dt}S(t\mid b,x) =-S(t\mid b,x)\sum_{j=1}^K\lambda_j(t\mid b,x), \qquad S(0\mid b,x)=1.

Solving this equation gives the conditional event-free survival function:

S(t\mid b,x) =\exp\left\{-\int_0^t\sum_{j=1}^K\lambda_j(u\mid b,x)\,du\right\}.

The conditional probability of a cause-$k$ event in a small interval is therefore weighted by the chance of surviving every cause up to that interval. Integrating a target-cause hazard without this weighting would count people as still at risk after a competing event could already have occurred.

Useful check: For mutually exclusive and exhaustive first-event causes, the conditional probabilities satisfy $$ S(t\mid b,x)+\sum_{k=1}^K F_k(t\mid b,x)=1. $$ Indeed, differentiating the left side gives $-S\sum_j\lambda_j+\sum_k S\lambda_k=0$, and its value at time zero is one. The same identity holds after averaging over random effects.

Worked Example: Constant Hazards for Two Competing Causes

Suppose, hypothetically, that for a fixed covariate profile and fixed random effects, the cause-1 hazard is $0.08$ per year and the cause-2 hazard is $0.12$ per year throughout follow-up. Calculate the cause-1 CIF at 3 years.

The total hazard is $0.08+0.12=0.20$ per year, so event-free survival at time $u$ is $S(u)=e^{-0.20u}$. The cause-1 CIF is

F_1(3)=\int_0^3 e^{-0.20u}(0.08)\,du =\frac{0.08}{0.20}\left(1-e^{-0.20(3)}\right).

Substituting and evaluating,

F_1(3)=0.4(1-e^{-0.6}) =0.4(1-0.5488) =0.1805.

Thus, under these hypothetical constant hazards, the probability of a cause-1 event by 3 years before a cause-2 event is about 18.1%. As a check, $S(3)=e^{-0.6}=0.5488$, while $F_2(3)=\frac{0.12}{0.20}(1-e^{-0.6})=0.2707$. Their sum with $F_1(3)$ is approximately 1.000.

Changing Hazards and Numerical Evaluation

When hazards vary over time, the same integral applies, but the event-free survival at each time must reflect the accumulated hazards of every cause up to that time. Piecewise-constant hazards make the calculation transparent: integrate separately over intervals, carrying forward the event-free survival at each interval boundary.

Worked Example: Piecewise-Constant Hazards

Consider hypothetical hazards over four years. During years 0 to 2, let $\lambda_1=0.10$ and $\lambda_2=0.20$ per year. During years 2 to 4, let $\lambda_1=0.20$ and $\lambda_2=0.10$ per year. Calculate the cause-1 CIF at year 4.

In the first interval, the total hazard is $0.30$ per year. The contribution to the CIF is

\int_0^2 e^{-0.30u}(0.10)\,du =\frac{0.10}{0.30}(1-e^{-0.60}) =0.1504.

Event-free survival at year 2 is $S(2)=e^{-0.60}=0.5488$. In the second interval, the total hazard is again $0.30$, and the cause-1 hazard is $0.20$. For $2\leq u\leq4$, survival is $0.5488e^{-0.30(u-2)}$. Therefore the second contribution is

\int_2^4 0.5488e^{-0.30(u-2)}(0.20)\,du =0.5488\frac{0.20}{0.30}(1-e^{-0.60}) =0.1651.

Adding the interval contributions gives $F_1(4)=0.1504+0.1651=0.3155$. The cause-1 event probability by year 4 is therefore about 31.6% under these hypothetical hazards. The calculation checks continuity of accumulated incidence at the interval boundary: the survival entering the second interval is the survival remaining after both causes operated in the first interval.

For hazards with no closed-form integral, evaluate $S(u\mid b,x)$ from the cumulative total hazard and numerically integrate $S(u\mid b,x)\lambda_k(u\mid b,x)$. A stable implementation should use a sufficiently fine time grid or a suitable numerical integration rule, and should verify that the resulting CIFs are nondecreasing, nonnegative, and no greater than one. The identity $S+\sum_k F_k=1$ is also a useful numerical check.

Averaging Over Random Effects

Conditional CIFs describe risk for a specified value of $b$. For a population with random-effects density $f(b\mid x)$, the marginal CIF is obtained by the law of total probability:

F_k(t\mid x) =\int F_k(t\mid b,x)f(b\mid x)\,db =\int\left\{\int_0^t S(u\mid b,x)\lambda_k(u\mid b,x)\,du\right\}f(b\mid x)\,db.

This averages conditional event probabilities, each calculated using the hazards and event-free survival for the same $b$. If the purpose is prediction for a particular person with longitudinal measurements, the relevant weights are instead the person-specific conditional distribution of $b$ given their observed information (including being event-free to the prediction time). The future CIF conditional on being event-free at $t_0$ uses the conditional survival ratio $S(u\mid b,x)/S(t_0\mid b,x)$ for $u\geq t_0$, and averages over that updated distribution of $b$.

Worked Example: Why the Random-Effects Average Matters

Suppose a hypothetical population has two equally common random-effects groups. In group A, the constant cause-specific hazards are $\lambda_1=0.10$ and $\lambda_2=0.10$ per year. In group B, they are $\lambda_1=0.30$ and $\lambda_2=0.10$ per year. Find the population cause-1 CIF at 2 years.

For group A, the total hazard is $0.20$, giving

F_{1,A}(2)=\frac{0.10}{0.20}(1-e^{-0.20(2)}) =0.5(1-e^{-0.4}) =0.1648.

For group B, the total hazard is $0.40$, giving

F_{1,B}(2)=\frac{0.30}{0.40}(1-e^{-0.40(2)}) =0.75(1-e^{-0.8}) =0.4130.

The population CIF is the equal-weight average of these conditional probabilities:

$$ F_1(2)=0.5(0.1648)+0.5(0.4130)=0.2889. $$

A tempting shortcut is to average the cause-specific hazards first, obtaining $\bar{\lambda}_1=0.20$ and $\bar{\lambda}_2=0.10$, then apply the constant-hazard formula. That gives $\frac{0.20}{0.30}(1-e^{-0.30(2)})=0.3008$, not $0.2889$. The shortcut fails because event-free survival depends nonlinearly on the total hazard. In general, integrate each conditional CIF over the random-effects distribution rather than plugging mean hazards into the conditional formula.

Interpretation, Alternatives, and Diagnostics

A CIF is an absolute probability in the presence of competing events. It is not the same quantity as a cause-specific hazard, which is an instantaneous rate among those still event-free, nor is it the probability obtained by pretending that competing causes cannot occur. Treating competing events as ordinary censoring and using a one-minus-survival calculation for the target cause generally produces a net-risk quantity rather than the observed-world CIF. For the CIF, other causes reduce the number who remain able to experience the target event.

The conditional and marginal formulations answer different questions. A conditional CIF fixes the random effects and covariates; a marginal CIF averages over a specified population distribution. A model-based CIF can be compared with nonparametric competing-risk estimates, such as the Aalen–Johansen estimator, which estimates event probabilities without specifying a parametric joint model. The model-based approach can use longitudinal information and covariate structure, but its validity depends on the hazard and random-effects models. Neither approach makes the CIF a causal effect without additional causal assumptions and a suitable study design.

Useful diagnostics include checking the fitted cause-specific hazard curves for implausible shapes, comparing predicted and observed event frequencies over time, and assessing calibration separately by cause and across relevant covariate or risk groups. For random-effects integration, check numerical stability by increasing the number of integration points or simulation draws; material changes in the CIF indicate inadequate integration. Also check that estimated cause-specific hazards remain nonnegative and that predicted $S(t)+\sum_kF_k(t)$ is close to one across the reported time range.

Specify the conditioning information.
State whether the target is conditional on fixed random effects or averaged over a population or patient-specific random-effects distribution.

Construct event-free survival.
Sum the cause-specific hazards and exponentiate the negative cumulative total hazard.

Integrate the target cause.
Integrate event-free survival times the cause-$k$ hazard from time zero to the target time.

Average and check.
If required, average conditional CIFs over the appropriate random-effects distribution, then check monotonicity and the probability-sum identity.

The main limitations follow from these requirements. Misspecified cause-specific hazards can distort both the target-cause contribution and survival from all causes. A misspecified random-effects distribution can distort population averaging, especially when hazards vary sharply with latent effects. Sparse events for one cause can make its hazard—and consequently its CIF—imprecise. Finally, an estimated CIF is a model-based probability for the specified covariate and random-effects distribution, not a guarantee for an individual.

Common Mistake

Wrong reasoning: “The cause-1 hazard is $0.08$ per year, so over three years the cause-1 event probability is $0.08\times3=0.24$.” This calculation ignores that people may have a competing event before cause 1 can occur. Correction: weight the target hazard at each time by survival from all causes, then integrate. With constant competing hazards, the relevant calculation is $\int_0^3 e^{-(\lambda_1+\lambda_2)u}\lambda_1\,du$, not $3\lambda_1$.

Check Your Understanding

Answer each question, then compare with the solutions below.

For constant cause-specific hazards $\lambda_1=0.05$ and $\lambda_2=0.15$ per year, calculate $F_1(2)$.
Why does the survival factor in the cause-$k$ CIF use the sum of hazards over all causes?
A population contains equal proportions of two random-effects groups. What quantity should be averaged to obtain the population CIF: the hazards or the conditional CIFs?
Name two checks that can help detect a computational or model problem in estimated CIFs.

Solutions. (1) The total hazard is $0.20$, so $F_1(2)=\frac{0.05}{0.20}(1-e^{-0.20(2)})=0.25(1-e^{-0.4})=0.08242$, or about 8.24% by 2 years. (2) An event of any cause removes the individual from the event-free risk set, so survival must account for every cause. (3) Average the conditional CIFs over the random-effects distribution; averaging hazards first generally gives a different result because survival depends nonlinearly on cumulative total hazard. (4) Suitable checks include nondecreasing CIFs bounded between zero and one, the identity $S+\sum_kF_k=1$, and stability as numerical integration accuracy is increased.